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Math Help - Integration by principles

  1. #1
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    Integration by principles

    Hello! Can someone please help me with this?

    I need to calculate the following integral by principle, that means resolving the limit of the sum.

    The integration segment [0,T] is partitioned in N partitions of size T/N.

    <br />
\int_0^{T} x^4\ dx\ = \lim_{N\to \infty}\sum_{i=0}^{N-1} x_{i}^{4}(x_{i+1} - x_{i})<br />

    Thanks a bunch!
    Last edited by paolopiace; March 3rd 2008 at 01:18 PM. Reason: clarity
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  2. #2
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    I suppose you're referring to a Riemann sum.

    {\Delta}x=\frac{T}{n}

    Using the right endpoint method, a+k{\Delta}x

    Since a=0, we have \frac{kT}{n}

    Sub into what is to be integrated:

    \left(\frac{kT}{n}\right)^{4}\cdot\frac{T}{n}

    \frac{t^{5}}{n^{5}}\sum_{k=1}^{n}k^{4}

    The thing to do here is to write the sum in terms of n.

    The sum of the first n fourth powers is given by:

    \frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}

    So, you get:

    \frac{t^{5}}{n^{5}}\left(\frac{n^{5}}{5}+\frac{n^{  4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}\right)

    Expand out and take the limit as n\rightarrow{\infty}

    You should get your answer the same as if you integrated the 'quick' way.
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  3. #3
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    Could be done easier?

    There must be some trick...

    For example:

    <br />
\int_0^{T} 2x\ dx\ = \lim_{N\to \infty}\sum_{i=0}^{N-1} 2x_{i}(x_{i+1} - x_{i}) = x^2(T) - T<br />

    That was done using the trick 2ab = [(a+b)^2 - a^2 - b^2] with a=x(i) and b=[x(i+1) - x(i)]

    The series becomes telescopic and yields the results...
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